ÌâÄ¿ÄÚÈÝ

ÉèNAΪ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£¬ÏÂÁÐÐðÊö´íÎóµÄÊÇ £¨  £©¡£
A£®±ê×¼×´¿öÏ£¬22.4 L SO2ºÍO2µÄ»ìºÏÆøÌåÖк¬ÓеÄÑõÔ­×ÓÊýΪ2NA
B£®±ê×¼×´¿öÏ£¬22.4 LÒÒÏ©Öк¬Óй²Óõç×Ó¶ÔµÄÊýĿΪ6NA
C£®ÃܱÕÈÝÆ÷ÖУ¬3 mol H2ºÍ1 mol N2ÔÚ¸ßΡ¢¸ßѹ¡¢´ß»¯¼ÁµÄÌõ¼þϳä·Ö·´Ó¦ºó£¬ÈÝÆ÷ÄÚµÄÆøÌå·Ö×Ó×ÜÊýΪ2NA
D£®2.8 gµÄCOºÍ2.8 gµÄN2Ëùº¬ÖÊ×ÓÊý¶¼Îª1.4NA
C
SO2ºÍO2·Ö×ÓÖж¼º¬Á½¸öÑõÔ­×Ó£¬AÕýÈ·£»ÓÉ1¸öÒÒÏ©·Ö×ÓÖк¬ÓÐ6¶Ô¹²Óõç×Ó¶Ô¿ÉÖª£¬BÕýÈ·£»ÒòH2ºÍN2ÔÚËù¸øÌõ¼þÏ·¢ÉúµÄ·´Ó¦ÊÇ¿ÉÄæ·´Ó¦£¬¹Ê3 mol H2ºÍ1 mol N2²»»áÍêÈ«·´Ó¦Éú³É2 mol NH3£¬¹ÊC´í£»COºÍN2µÄĦ¶ûÖÊÁ¿ÏàµÈ£¬¾ùΪ28 g¡¤mol£­1,2.8 g COºÍ2.8 g N2Ëùº¬ÖÊ×ÓÊýÒ²ÏàµÈ£¬¾ùΪ1.4NA£¬DÕýÈ·¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
µªÊǵØÇòÉϺ¬Á¿·á¸»µÄÒ»ÖÖÔªËØ£¬°±¡¢ëÂ(N2H4)ºÍµþµªËᶼÊǵªÔªËصÄÖØÒªÇ⻯ÎÔÚ¹¤Å©ÒµÉú²ú¡¢Éú»îÖÐÓÐ×ÅÖØ´ó×÷Óá£
(1)ºÏ³É°±Éú²ú¼¼ÊõµÄ´´Á¢¿ª±ÙÁËÈ˹¤¹ÌµªµÄ;¾¶£¬¶Ô»¯Ñ§¹¤Òµ¼¼ÊõÒ²²úÉúÁËÖØÒªÓ°Ïì¡£
¢ÙÔڹ̶¨Ìå»ýµÄÃܱÕÈÝÆ÷ÖУ¬½øÐÐÈçÏ»¯Ñ§·´Ó¦£ºN2(g)£«3H2(g)2NH3(g)¡¡¦¤H£¼0£¬Æäƽºâ³£ÊýKÓëζÈTµÄ¹ØϵÈçÏÂ±í¡£
T/K
298
398
498
ƽºâ³£ÊýK
4.1¡Á106
K1
K2
 
Ôò¸Ã·´Ó¦µÄƽºâ³£ÊýµÄ±í´ïʽΪ________£»ÅжÏK1________K2(Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±)¡£
¢ÚÏÂÁи÷ÏîÄÜ˵Ã÷¸Ã·´Ó¦ÒѴﵽƽºâ״̬µÄÊÇ________(Ìî×Öĸ)¡£
a£®ÈÝÆ÷ÄÚN2¡¢H2¡¢NH3µÄŨ¶ÈÖ®±ÈΪ1¡Ã3¡Ã2
b£®v(N2)Õý£½3v(H2)Äæ
c£®ÈÝÆ÷ÄÚѹǿ±£³Ö²»±ä
d£®»ìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä
¢ÛÒ»¶¨Î¶ÈÏ£¬ÔÚ1 LÃܱÕÈÝÆ÷ÖгäÈë1 mol N2ºÍ3 mol H2²¢·¢ÉúÉÏÊö·´Ó¦¡£ÈôÈÝÆ÷ÈÝ»ýºã¶¨£¬10 min´ïµ½Æ½ºâʱ£¬ÆøÌåµÄ×ÜÎïÖʵÄÁ¿ÎªÔ­À´µÄ£¬ÔòN2µÄת»¯ÂÊΪ________£¬ÒÔNH3µÄŨ¶È±ä»¯±íʾ¸Ã¹ý³ÌµÄ·´Ó¦ËÙÂÊΪ________¡£
(2)ë¿ÉÓÃÓÚ»ð¼ýȼÁÏ¡¢ÖÆÒ©Ô­Áϵȡ£
¢ÙÔÚ»ð¼ýÍƽøÆ÷ÖÐ×°ÓÐëÂ(N2H4)ºÍҺ̬H2O2£¬ÒÑÖª0.4 molҺ̬N2H4ºÍ×ãÁ¿ÒºÌ¬H2O2·´Ó¦£¬Éú³ÉÆø̬N2ºÍÆø̬H2O£¬·Å³ö256.6 kJµÄÈÈÁ¿¡£¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ________________________________________________________________________¡£
¢ÚÒ»ÖÖëÂȼÁϵç³ØµÄ¹¤×÷Ô­ÀíÈçͼËùʾ¡£¸Ãµç³Ø¹¤×÷ʱ¸º¼«µÄµç¼«·´Ó¦Ê½Îª_____________________________________¡£

¢Û¼ÓÈÈÌõ¼þÏÂÓÃҺ̬ëÂ(N2H4)»¹Ô­ÐÂÖÆCu(OH)2À´ÖƱ¸ÄÉÃ×¼¶Cu2O£¬Í¬Ê±·Å³öN2¡£¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ_______________________________________¡£
ëÂÓëÑÇÏõËá(HNO2)·´Ó¦¿ÉÉú³ÉµþµªËᣬ8.6 gµþµªËáÍêÈ«·Ö½â    ¿É·Å³ö6.72 LµªÆø(±ê×¼×´¿öÏÂ)£¬ÔòµþµªËáµÄ·Ö×ÓʽΪ________¡£
(1)ÏÂÁÐʵÑé²Ù×÷»ò¶ÔʵÑéÊÂʵµÄÃèÊö²»ÕýÈ·µÄÊÇ________(ÌîÐòºÅ)¡£
A£®²»ÒËÓôÉÛáÛö×ÆÉÕÇâÑõ»¯ÄÆ»ò̼ËáÄÆ
B£®Ê¹ÓÃÈÝÁ¿Æ¿ÅäÖÆÈÜҺʱ£¬¸©Êӿ̶ÈÏ߶¨ÈݺóËùµÃÈÜҺŨ¶ÈÆ«´ó
C£®·ÖÒº²Ù×÷ʱ£¬ÓÉÓÚ·ÖҺ©¶·Ï¶ËÒÑÕ´ÂúϲãÒºÌ壬¹ÊÉϲãÒºÌåÒª´ÓÉÏ¿Úµ¹³ö
D£®ÓÃÍÐÅÌÌìƽ³ÆÁ¿11.74 gÂÈ»¯Äƾ§Ìå
E£®Óýᾧ·¨·ÖÀëÏõËá¼ØºÍÂÈ»¯ÄƵĻìºÏÎï
(2)ʵÑéÊÒÐèÒª0.1 mol¡¤L£­1 NaOHÈÜÒº450 mLºÍ0.5 mol¡¤L£­1ÁòËáÈÜÒº500 mL¡£¸ù¾ÝÕâÁ½ÖÖÈÜÒºµÄÅäÖÆÇé¿ö»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÈçͼËùʾµÄÒÇÆ÷ÖÐÅäÖÆÈÜÒº¿Ï¶¨²»ÐèÒªµÄÊÇ________(ÌîÐòºÅ)£»ÅäÖÆÉÏÊöÈÜÒº»¹ÐèÓõ½µÄ²£Á§ÒÇÆ÷ÊÇ________(ÌîÒÇÆ÷Ãû³Æ)¡£

¢ÚÏÂÁвÙ×÷ÖУ¬ÈÝÁ¿Æ¿Ëù²»¾ß±¸µÄ¹¦ÄÜÓÐ________(ÌîÐòºÅ)¡£
A£®ÅäÖÆÒ»¶¨Ìå»ý׼ȷŨ¶ÈµÄ±ê×¼ÈÜÒº
B£®Öü´æÈÜÒº
C£®²âÁ¿ÈÝÁ¿Æ¿¹æ¸ñÒÔϵÄÈÎÒâÌå»ýµÄÒºÌå
D£®×¼È·Ï¡ÊÍijһŨ¶ÈµÄÈÜÒº
E£®ÓÃÀ´¼ÓÈÈÈܽâ¹ÌÌåÈÜÖÊ
¢ÛÇë¼ÆËãÐèÓÃÍÐÅÌÌìƽ³ÆÈ¡¹ÌÌåNaOHµÄÖÊÁ¿Îª________g¡£Ä³Í¬Ñ§ÓÃÍÐÅÌÌìƽ³ÆÁ¿ÉÕ±­µÄÖÊÁ¿£¬ÌìƽƽºâºóµÄ״̬Èçͼ¡£ÓÉͼÖпÉÒÔ¿´³ö£¬ÉÕ±­µÄʵ¼ÊÖÊÁ¿Îª________g¡£

¢Ü¸ù¾Ý¼ÆËãµÃÖª£¬ËùÐèÖÊÁ¿·ÖÊýΪ98%¡¢ÃܶÈΪ1.84 g¡¤cm£­3µÄŨÁòËáµÄÌå»ýΪ________mL(¼ÆËã½á¹û±£ÁôһλСÊý)¡£ÅäÖƹý³ÌÖÐÐèÏÈÔÚÉÕ±­Öн«Å¨ÁòËá½øÐÐÏ¡ÊÍ£¬Ï¡ÊÍʱ²Ù×÷·½·¨ÊÇ_____________________________________________¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø